Question

Find the values of a and b so that the function

\[f\left( x \right)\begin{cases}x^2 + 3x + a, & \text { if } x \leq 1 \\ bx + 2 , &\text {  if } x > 1\end{cases}\] is differentiable at each x ∈ R.

Answer 1

Given: 

\[f(x) = \binom{ x^2 + 3x + a, x \leq 1}{bx + 2, x > 1}\]

It is given that the function is differentiable at each 

\[x \in R\]  and every differentiable function is continuous.

So, 

\[f(x)\]  is continuous at

\[x = 1\]

Therefore,

\[\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)\]

\[\Rightarrow \lim_{x \to 1} x^2 + 3x + a = \lim_{x \to 1} bx + 2 = a + 4 \left[ \text { Using def . of } f(x) \right]\]
\[ \Rightarrow a + 4 = b + 2 = a + 4 . . . (i)\]

Since,  

\[f(x)\]  is differentiable at 

\[x = 1\] . So, 

(LHD at x = 1) = (RHD at x = 1)
\[\lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^+} \frac{f(x) - f(1)}{x - 1}\]
\[ \Rightarrow \lim_{x \to 1} \frac{x^2 + 3x + a - a - 4}{x - 1} = \lim_{x \to 1} \frac{bx + 2 - 4 - a}{x - 1} \left[ \text { Using def . of } f(x) \right]\]
\[ \Rightarrow \lim_{x \to 1} \frac{(x + 4) (x - 1)}{x - 1} = \lim_{x \to 1} \frac{bx - 2 - a}{x - 1}\]
\[ \Rightarrow \lim_{x \to 1} \frac{(x + 4) (x - 1)}{x - 1} = \lim_{x \to 1} \frac{bx - b}{x - 1} \left[ \text { Using } (i) \right] \]
\[ \Rightarrow \lim_{x \to 1} \frac{(x + 4) (x - 1)}{x - 1} = \lim_{x \to 1} \frac{b(x - 1)}{x - 1}\]
\[ \Rightarrow 5 = b\]

From  

\[(i)\], we have

\[a + 4 = b + 2\]
\[ \Rightarrow a + 4 = 5 + 2\]
\[ \Rightarrow a = 7 - 4 \]
\[ \Rightarrow a = 3\]

Hence, 

\[a = 3 , b = 5\].